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| Derivation or Formula | Reasoning |
|---|---|
| \[ \Delta x = 6.2\cos(1.6\,t) \] | This is the standard displacement equation for simple harmonic motion, with an amplitude of \(6.2\,m\) and a phase constant of zero since the object is at its maximum displacement at \(t = 0.0\,s\). |
| \[ \Delta x = 6.2\cos(1.6 \times 3.5) \] | We substitute the given time \(t = 3.5\,s\) into the equation. |
| \[ 1.6 \times 3.5 = 5.6 \] | This multiplication gives the phase angle in radians used in the cosine function. |
| \[ \cos(5.6) \approx 0.777 \] | The cosine of \(5.6\,rad\) is approximately \(0.777\) (using a calculator or estimation by relating it to \(2\pi – 5.6\)). |
| \[ \Delta x \approx 6.2 \times 0.777 \approx 4.8\,\text{m} \] | Multiplying the amplitude by the cosine value gives the displacement at \(t = 3.5\,s\). |
| \[ \boxed{4.8\,\text{m}} \] | This is the final displacement of the object, which matches option (a). |
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Block \( 1 \) of mass \( m_1 \) and Block \( 2 \) of mass \( m_2 = 2 m_1 \) are each attached to identical horizontal springs. Each block is displaced from equilibrium by an unknown amount and the blocks are released from rest simultaneously, undergoing simple harmonic motion. A student claims that Block \( 1 \) will make its first return to its equilibrium position before Block \( 2 \) first returns to its equilibrium position. Is this claim correct? Why or why not?
When the mass of a simple pendulum is tripled, the time required for one complete vibration
A \( 50 \) \( \text{g} \) ice cube can slide up and down a frictionless \( 30^{\circ}\) slope. At the bottom, a spring with spring constant \( 25 \) \( \text{N/m} \) is compressed \( 10 \) \( \text{cm} \) and is used to launch the ice cube up the slope. How high does it go above its starting point? Express your answer with the appropriate units.
An experimenter has a simple pendulum of length \( L \) and a mass–spring system with mass \( m \) and spring constant \( k \). Both are found to have the same period of oscillation \( T \) on Earth. If both systems are taken to the Moon, where the acceleration due to gravity is approximately \( \frac{1}{6} g \) of Earth, what will happen to their periods?
Students attach a thin strip of metal to a table so that the strip is horizontal in relation to the ground. A section of the strip hangs off the edge of the table. A mass is secured to the end of the hanging section of the strip and is then displaced so that the mass-strip system oscillates, as shown in the figure. Students make various measurements of the net force F exerted on the mass as a result of the force due to gravity and the normal force from the strip, the vertical position y of the mass above and below its equilibrium position y. and the period of oscillation T’ when the mass is displaced by different amplitudes A. Which of the following explanations is correct about the evidence required to conclude that the mass undergoes simple harmonic motion?
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
A spring with a spring constant of \( 600. \) \( \text{N/m} \) is used for a scale to weigh fish. What is the mass of a fish that would stretch the spring by \( 7.5 \) \( \text{cm} \) from its normal length?
The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
A linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
A bullet (mass: \(0.05 \, \text{kg}\)) is fired horizontally (\(v = 200 \, \text{m/s}\)) at a block (mass: \(1.3 \, \text{kg}\)) initially at rest on a frictionless surface. The block is attached to a spring (\(k = 2500 \, \text{N/m}\)). The bullet becomes embedded. Calculate:
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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